B

Step 1: The range of a projectile is given by the formula:
$$ R = \frac{v^2 \sin(2\theta)}{g} $$
where v is the initial velocity, g is the acceleration due to gravity, and \theta is the angle of projection.

Step 2: Given the range at 75º is 0.5 km, we can express it as:
$$ R_{75} = \frac{v^2 \sin(150^\circ)}{g} = 0.5 $$
and since $\sin(150^\circ) = \frac{1}{2}$,
$$ R_{75} = \frac{v^2 \cdot \frac{1}{2}}{g} = 0.5 $$
This simplifies to:
$$ \frac{v^2}{2g} = 0.5 $$
Therefore, we have:
$$ v^2 = g \cdot 1 $$

Step 3: Now, let's calculate the range when fired at 45º:
Using the range formula again:
$$ R_{45} = \frac{v^2 \sin(90^\circ)}{g} = \frac{v^2}{g} $$

Step 4: Substitute for $v^2$ from our previous calculation:
$$ R_{45} = \frac{g \cdot 1}{g} = 1.0 $$

Therefore, the range when fired at 45º with the same speed is 1.0 km.

Hence, the correct answer is Option B.